CHAPTER 11
SIZE REDUCTION
Raw materials often occur in sizes that are too large to be used and, therefore, they must be
reduced in size. This size reduction operation can be divided into two major categories
depending on whether the material is a solid or a liquid. If it is solid, the operations are called
grinding and cutting, if it is liquid, emulsification or atomization. All depend on the reaction
to shearing forces within solids and liquids.
GRINDING AND CUTTING
Grinding and cutting reduce the size of solid materials by mechanical action, dividing them
into smaller particles. Perhaps the most extensive application of grinding in the food industry
is in the milling of grains to make flour. But it is used in many other processes, such as in the
grinding of corn for manufacture of corn starch, the grinding of sugar and the milling of dried
foods, such as vegetables. Cutting is used to break down large pieces of food into smaller
pieces suitable for further processing, such as in the preparation of meat for retail sales and in
the preparation of processed meats and processed vegetables.
In the grinding process, materials are reduced in size by fracturing them. The mechanism of
fracture is not fully understood, but in the process, the material is stressed by the action of
mechanical parts of the grinding machine and initially the stress is absorbed internally by the
material as strain energy. When the local strain energy exceeds a critical level, which is a
function of the material, fracture occurs along lines of weakness and the stored energy is
released. Some of the energy is taken up in the creation of new surface, but the greater part of
it is dissipated as heat.
Time also plays a part in the fracturing process and it appears that material will fracture at
lower stress concentrations if these can be maintained for longer periods. Grinding is,
therefore, achieved by mechanical stress followed by rupture and the energy required
depends upon the hardness of the material and also upon the tendency of the material to crack
- its friability.
The force applied may be compression, impact, or shear, and both the magnitude of the force
and the time of application affect the extent of grinding achieved. For efficient grinding, the
energy applied to the material should exceed, by as small a margin as possible, the minimum
energy needed to rupture the material. Excess energy is lost as heat and this loss should be
kept as low as practicable.
The important factors to be studied in the grinding process are the amount of energy used and
the amount of new surface formed by grinding.
Energy Used in Grinding
Grinding is a very inefficient process and it is important to use energy as efficiently as
possible. Unfortunately, it is not easy to calculate the minimum energy required for a given
reduction process, but some theories have been advanced which are useful.
These theories depend upon the basic assumption that the energy required to produce a
change dL in a particle of a typical size dimension L is a simple power function of L:
dE/dL = KLn
(11.1)
where dE is the differential energy required, dL is the change in a typical dimension, L is the
magnitude of a typical length dimension and K, n, are constants.
Kick assumed that the energy required to reduce a material in size was directly proportional
to the size reduction ratio dL/L. This implies that n in eqn. (11.1) is equal to -1. If
K = KKfc
where KK is called Kick's constant and fc is called the crushing strength of the material, we
have:
dE/dL = KKfcL-1
which, on integration gives:
E = KKfc ln(L1/L2)
(11.2)
Equation (11.2) is a statement of Kick's Law. It implies that the specific energy required to
crush a material, for example from 10cm down to 5cm, is the same as the energy required to
crush the same material from 5mm to 2.5 mm.
Rittinger, on the other hand, assumed that the energy required for size reduction is directly
proportional, not to the change in length dimensions, but to the change in surface area. This
leads to a value of -2 for n in eqn. (11.1, as area is proportional to length squared. If we put:
K = K R fc
And so;
dE/dL = KRfcL-2
where KR is called Rittinger's constant, and integrate the resulting form of eqn. (11.1), we
obtain:
E = KRfc(1/L2 – 1/L1)
(11.3)
Equation (11.3) is known as Rittinger's Law. As the specific surface of a particle (the surface
area per unit mass) is proportional to 1/L, eqn. (11.3) postulates that the energy required to
reduce L for a mass of particles from 10cm to 5cm would be the same as that required to
reduce, for example, the same mass of 5mm particles down to 4.7mm. This is a very much
smaller reduction, in terms of of energy per unit mass for the smaller particles, than that
predicted by Kick's Law.
It has been found, experimentally, that for the grinding of coarse particles in which the
increase in surface area per unit mass is relatively small, Kick's Law is a reasonable
approximation. For the size reduction of fine powders, on the other hand, in which large areas
of new surface are being created, Rittinger's Law fits the experimental data better.
Bond has suggested an intermediate course, in which he postulates that n is (-3/2) and this
leads to:
E = Ei (100/L2) 1/2[1 – (1/q) 1/2]
(11.4)
Bond defines the quantity Ei by this equation. L is measured in microns in eqn. (11.4) and so
Ei is the amount of energy required to reduce unit mass of the material from an infinitely
large particle size down to a particle size of 100m. It is expressed in terms of q, the
reduction ratio where q = L1/L2.
Note that all of these equations [eqns. (11.2), (11.3), and (11.4)] are dimensional equations
and so if quoted values are to be used for the various constants, the dimensions must be
expressed in appropriate units. In Bond's equation, if L is expressed in microns, this defines
Ei and Bond calls this the Work Index.
The greatest use of these equations is in making comparisons between power requirements
for various degrees of reduction.
EXAMPLE 11.1. Grinding of sugar
Sugar is ground from crystals of which it is acceptable that 80% pass a 500m sieve (US
Sieve No.35), down to a size in which it is acceptable that 80% passes a 88m sieve
(No.170), and a 5horsepower motor is found just sufficient for the required throughput. If the
requirements are changed such that the grinding is only down to 80% through a 125m sieve
(No 120) but the throughput is to be increased by 50% would the existing motor have
sufficient power to operate the grinder? Assume Bond's equation.
Using the subscripts 1 for the first condition and 2 for the second, and letting m kgh-1 be the
initial throughput, then if x is the required power
88m = 88x10-6m, 125m = 125x10-6m, 500m= 500x10-6m
= Ei(100/88 x 10-6)1/2 [1 – (88/500)1/2 ]
E1
= 5/m
E2
= x/1.5m
E2/E1
= x/(1.5 x 5) = (88 x 10-6)1/2 [1 - (125/500)1/2]
(125 x 10-6)1/2 [1 – (88/500)1/2]
= Ei(100/125 x 10-6)1/2 [1 - (125/500)1/2 ]
x/(7.5) = 0.84 x (0.500/0.58)
= 0.72
x
= 5.4 horsepower.
So the motor would be expected to have insufficient power to pass the 50% increased
throughput, though it should be able to handle an increase of 40%.
New Surface Formed by Grinding
When a uniform particle is crushed, after the first crushing the size of the particles produced
will vary a great deal from relatively coarse to fine and even to dust. As the grinding
continues, the coarser particles will be further reduced but there will be less change in the
size of the fine particles. Careful analysis has shown that there tends to be a certain size that
increases in its relative proportions in the mixture and which soon becomes the predominant
size fraction. For example, wheat after first crushing gives a wide range of particle sizes in
the coarse flour, but after further grinding the predominant fraction soon becomes that
passing a 250m sieve and being retained on a 125m sieve. This fraction tends to build up,
however long the grinding continues, so long as the same type of machinery, rolls in this
case, is employed.
The surface area of a fine particulate material is large and can be important. Most reactions
are related to the surface area available, so the surface area can have a considerable bearing
on the properties of the material. For example, wheat in the form of grains is relatively stable
so long as it is kept dry, but if ground to a fine flour has such a large surface per unit mass
that it becomes liable to explosive oxidation, as is all too well known in the milling industry.
The surface area per unit mass is called the specific surface. To calculate this in a known
mass of material it is necessary to know both the particle size distribution and also the shape
factor of the particles. The particle size gives one dimension that can be called the typical
dimension, Dp, of a particle. This has now to be related to the surface area.
We can write, arbitrarily:
Vp = pDp3
and
Ap= 6qDp2
where Vp is the volume of the particle, Ap is the area of the particle surface, Dp is the typical
dimension of the particle and p, q are factors which connect the particle geometries.(Note
subscript p and factor p.)
For example, for a cube, the volume is Dp3 and the surface area is 6Dp2; for a sphere the
volume is (/6)Dp3 and the surface area is Dp2 In each case the ratio of surface area to
volume is 6/Dp.
A shape factor is now defined as q/p = , (lambda), so that for a cube or a sphere  = 1. It has
been found, experimentally, that for many materials when ground, the shape factor of the
resulting particles is approximately 1.75, which means that their surface area to volume ratio
is nearly twice that for a cube or a sphere.
The ratio of surface area to volume is:
and so
Ap /Vp = 6q/pDp = 6/Dp
Ap = 6q Vp /pDp
= 6Vp /Dp
(11.5)
If there is a mass m of particles of density p the number of particles is m/pVp each of area Ap
So total area
At = (m/pVp)( 6qVp/pDp) = (6qm/p pDp)
= 6m/pDp
(11.6)
where At is the total area of the mass of particles. Equation (11.6) can be combined with the
results of sieve analysis to estimate the total surface area of a powder.
EXAMPLE 11.2. Surface area of salt crystals
In an analysis of ground salt using Tyler sieves, it was found that 38%of the total salt passed
through a 7mesh sieve and was caught on a 9mesh sieve. For one of the finer fractions, 5%
passed an 80mesh sieve but was retained on a 115mesh sieve. Estimate the surface areas of
these two fractions in a 5kg sample of the salt, if the density of salt is 1050kgm -3 and the
shape factor ( ) is 1.75.
Aperture of Tyler sieves, 7 mesh = 2.83mm, 9 mesh = 2.00mm, 80 mesh = 0.177mm, 115
mesh = 0.125mm.
Mean aperture 7 and 9 mesh = 2.41mm = 2.41 x10-3m
Mean aperture 80 and 115 mesh = 0.151 mm = 0.151 x10-3m.
Now from equn.(11.6)
A1
= (6 x 1.75 x 0.38 x 5)/(1050 x 2.41 x 10-3)
= 7.88m2
A2
= (6 x 1.75 x 0.05 x 5)/(1050 x 0.151 x 10-3)
= 16.6m2
Grinding Equipment
Grinding equipment can be divided into two classes - crushers and grinders. In the first class
the major action is compressive, whereas grinders combine shear and impact with
compressive forces.
Crushers
Jaw and gyratory crushers are heavy equipment and are not used extensively in the food
industry.
In the jaw crusher, the material is feed between two heavy jaws, one fixed and the other
reciprocating, so as to work the material down into a narrower and narrower space, crushing
it as it goes.
The gyratory crusher consists of a truncated conical casing, inside which a crushing head
rotates eccentrically. The crushing head is shaped as an inverted cone and the material being
crushed is trapped between the outer fixed, and the inner gyrating, cones, and it is again
forced into a narrower and narrower space during which time it is crushed.
Jaw and gyratory crusher actions are illustrated in Fig 11.1 (a) and (b).
Figure 11.1 Crushers: (a) jaw (b) gyratory
Crushing rolls consist of two heavy horizontal cylinders, mounted parallel to each other and
close together. They rotate in the opposite direction and the material to be crushed is trapped
and nipped between them, being crushed as it passes through. In some case, the rolls are both
driven at the same speed. In other cases, they may be driven at differential speeds, or only
one roll is driven. A major application is in the cane sugar industry, where several stages of
rolls are used to crush the cane
Hammer mills
In a hammer mill, swinging hammerheads are attached to a rotor that rotates at high speed
inside a hardened casing. The principle is illustrated in Fig. 11.2(a).
Figure 11.2 Grinders: (a) hammer mill, (b) plate mill
The material is crushed and pulverized between the hammers and the casing and remains in
the mill until it is fine enough to pass through a screen which forms the bottom of the casing.
Both brittle and fibrous materials can be handled in hammer mills, though with fibrous
material, projecting sections on the casing may be used to give a cutting action.
Plate mills
In plate mills the material is fed between two circular plates, one of them fixed and the other
rotating. The feed comes in near the axis of rotation and is sheared and crushed as it makes its
way to the edge of the plates, see Fig. 11.2(b). The plates can be mounted horizontally as in
the traditional Buhr stone used for grinding corn, which has a fluted surface on the plates.
The plates can be mounted vertically also. Developments of the plate mill have led to the
colloid mill, which uses very fine clearances and very high speeds to produce particles of
colloidal dimensions.
Fixed head mills
Various forms of mills are used in which the material is sheared between a fixed casing and a
rotating head, often with only fine clearances between them. One type is a pin mill in which
both the static and the moving plates have pins attached on the surface and the powder is
sheared between the pins.
Roller mills
Roller mills are similar to roller crushers, but they have smooth or finely fluted rolls, and
rotate at differential speeds. They are used very widely to grind flour. Because of their simple
geometry, the maximum size of the particle that can pass between the rolls can be regulated.
If the friction coefficient between the rolls and the feed material is known, the largest particle
that will be nipped between the rolls can be calculated, knowing the geometry of the
particles.
Miscellaneous milling equipment
The range of milling equipment is very wide. It includes ball mills, in which the material to
be ground is enclosed in a horizontal cylinder or a cone and tumbled with a large number of
steel balls, natural pebbles or artificial stones, which crush and break the material. Ball mills
have limited applications in the food industry, but they are used for grinding food colouring
materials. The edge-runner mill, which is basically a heavy broad wheel running round a
circular trough, is used for grinding chocolate and confectionery. Many types of milling
equipment have come to be traditional in various industries and it is often claimed that they
provide characteristic actions that are peculiarly suited to, and necessary for, the product.
Cutters
Cutting machinery is generally simple, consisting of rotating knives in various arrangements.
A major problem often is to keep the knives sharp so that they cut rather than tear. An
example is the bowl chopper in which a flat bowl containing the material revolves beneath a
vertical rotating cutting knife.
EMULSIFICATION
Emulsions are stable suspensions of one liquid in another, the liquids being immiscible.
Stability of the emulsion is obtained by dispersion of very fine droplets of one liquid, called
the disperse phase, through the other liquid, which is called the continuous phase. The
emulsion is stable when it can persist without change, for long periods of time, without the
droplets of the disperse phase coalescing with each other, or rising or settling. The stability of
an emulsion is controlled by
• interfacial surface forces,
• size of the disperse phase droplets,
• viscous properties of the continuous phase and
• density difference between the two phases.
The dispersed particles in the emulsion have a very large surface area, which is created in the
process of emulsification. Surface effects depend upon the properties of the materials of the
two phases, but very often a third component is added which is absorbed at the interface and
which helps to prevent the droplets from coalescing. These added materials are called
emulsifying agents and examples are phosphates and glycerol monostearate.
The size of the disperse phase droplets is important and these are commonly of the order 1-10
m diameter. Below 0.1m droplet diameter, the dispersion is often spoken of as colloidal.
Coalescence of the disperse phase droplets is hindered by increased viscosity in the
continuous liquid phase. The nearer the densities of the components are to each other, the less
will be the separating effect of gravitational forces. Stokes' Law gives a qualitative indication
of the physical factors that influence the stability of an emulsion. This is because the relative
flow of the particles under gravitational forces may break the emulsion, so stability is
enhanced by small settling velocities. From eqn. (10.2):
Vm = D2g(p - f )/18
the critical importance of particle size, occurring as a squared term, can be seen. Also it
shows why emulsions are more stable when density differences are small and when the
viscosity of the continuous phase is high.
Preparation of Emulsions
The essential feature of an emulsion is the small size of the disperse phase droplets. This can
be achieved by imposing very high shearing stresses upon the liquid that is to be dispersed
and the shearing forces break the material into the multitude of fine particles.
Shearing is, generally, attained by passing the liquid through a high pressure pump, to bring it
up to pressures of the order of 7 x 103kPa, and then discharging this pressure suddenly by
expansion of the liquid through a small gap or nozzle; the equipment is often called a
homogenizer. In passing through the nozzle, very large shear forces are exerted on the liquid,
disrupting cohesion and dispersing it into the very small particles.
Centrifugal orifices may also be used to obtain the shearing action. Discs spinning at high
velocities give rise to high shearing forces in liquids flowing over them. Flow between
contra-rotating discs, which may have pegs on the disc faces, can be used to produce
emulsions. Designs in which small clearances are used between a stationary disc and a high
speed flat or conical rotating disc are called colloid mills. Another source of energy for
shearing is from ultrasonic vibrations induced in the liquid.
Existing emulsions can be given increased stability by decreasing the size of the droplets
either by impact or shearing the emulsion still further; the process is called homogenization.
Homogenizing results in smaller and more uniform droplet sizes and a practical example is
the homogenizing of milk.
Examples of emulsions met with frequently in the food industry are - milk (fat dispersed in
water), butter (water dispersed in fat), mayonnaise (oil in water) and ice cream (fat in water
which is then frozen).
Milk is an emulsion of fat in water, which is not stable indefinitely as it separates on
standing, into skim milk and cream. This is caused by the density differences between the fat
and the water, the fat globules rising as predicted by Stokes' Law and coalescing at the
surface to form a layer of cream. After homogenizing, this separation does not occur as the
globules are much reduced in size. Homogenizing is also used with ice cream mixes, which
are dispersions of fat and air in sugar solutions, and in the manufacture of margarine.
The same surface effects that govern liquid emulsions also apply to dispersions of solids in
liquids and of liquids or solids in gases. Colloidal solutions of solids can be produced if the
particle size is of the necessary order, below about 0.1m, and again stability depends upon
the surface properties of the materials. Aerosols, for example, fine mists in the atmosphere,
can also be quite stable.
SUMMARY
1. Size reduction is accomplished by shearing forces that cause the material to fracture
releasing most of the applied energy as heat.
2. A general equation giving the power required for size reduction is:
dE/dL = KLn
and from this can be derived
(a) Kick's Law in which n = 1 and which may be integrated to give:
E = KKfc ln(L1/L2)
(b) Rittinger's Law in which n = -2, integrated to give:
E = KR fc(1/L2 – 1/L1)
(c) Bond's equation in which n = - 3/2, integrated to give:
E = Ei (100/L2)1/2[1 – (1/q)1/2]
It appears that Kick's results apply better to coarser particles, Rittinger's to fine ones with
Bond's being intermediate.
3. The total surface area of a powder is important and can be estimated from
At = 6m/pDp
4. An emulsion is produced by shearing forces, which reduce the size of droplets of the
dispersed phase to diameters of the order of 0.1-10m, with a large specific surface area.
Application of Stokes' Law gives an indication of emulsion stability.
PROBLEMS
1. From measurements on a uniformly sized material from a dryer, it is inferred that the
surface area of the material is 1200m2. If the density of the material is 1450kgm-3 and the
total weight is 360kg, calculate the equivalent diameter of the particles if their value of 
is 1.75.
(2200microns)
2. Calculate the shape factor (  ) for model systems in which the particles are:
(a) cylinders with L = 2D,
(b) tetrahedra with their sides being equilateral triangles; the volume of a
tetrahedron being the area of the base x 1/3 of the vertical height.
(c) Estimate the specific surface area of a powder consisting of equal numbers, of equal
total weight, of the above two shapes in which there are 4 x 103particles kg-1. The cylinders
have a density of 1330kgm-3 and the tetrahedra a density of 1500kgm-3.
((a) 0.83, (b) 2.4 (c) 0.81m2 kg-1)
3. It is found that the energy required to reduce particles from a mean diameter of 1cm to
0.3cm is 11 kJ kg-1. Estimate the energy requirement to reduce the same particles from a
diameter of 0.1 cm to 0.01 cm assuming:
(a) Kick's Law,
(b) Rittinger's Law,
(c) Bond's Equation.
((a) 21kJkg-1 (b) 423 kJkg-1 (c) 91 kJkg-1)
4. It is suspected that for a product of interest the oxidation reactions, which create offflavours, are surface reactions, which proceed at a rate that is uniform with time. If the
shelf life of the product is directly related to the percentage of the off-flavours that have
been produced, estimate the reduction in shelf life consequent upon the size reductions of
example 3, that is (a) from 1cm to 0.3cm and (b) from 0.1cm to 0.01cm in diameter,
assuming  = 1.5.
((a) 10:1 (b) 100:1)
5. If it is desired to reduce the separation time for milk to at least one week (before cream
will rise to the top), what maximum diameter of cream droplet would Stokes' Law predict
to be necessary for the homogenization to achieve?
(0.0567 microns)